# Definition talk:Product Space (Topology)

## Tychonoff Topology

Let $\mathcal{T}$ be the Tychonoff topology on $X$. Since $\mathcal{T}$ is the topology generated by $\mathcal S = \left\{{\operatorname {pr}_i^{-1} \left({U}\right) : i \in I, U \in \vartheta_i}\right\}$ isn't it also the topology that renders all projections **continuous** (linear) functionals? Therefore, $\mathcal{T}$ is the product topology on $X$. What one needs to prove is that the functions:

- $\pi_1(x)=x_1$

and

- $\pi_2(x)=x_2$

where $x\in X$ is $x=(x_1,x_2)$ with $x_1\in X_1$ and $x_2\in X_2$,

are continuous with respect to the described topology with basis $\mathcal{B}$. However, I don't like very much this definition because firstly it refers only to cartesian products between two topological spaces (while you can have Cartesian products made up of arbitrarily many such spaces) and because most textbooks provide the definition that the product topology is the topology that renders all projections continuous (i.e. exactly this definition). For example S. Axler and K.A. Ribet, "A Taste of Topoology", Springer Editions, Berlin 2005, ISBN: 0-387-25790-X. I suggest that we modify the definition of Product topology and have just a link to the Tychonoff topology.

- While I agree with most of what you say, some remarks are in order. First of all, the projections aren't functionals in the functional analytic sense of the word (which is the only one known to me), they are operators. Linearity subsumes an additive structure, which isn't given. Lastly, please be aware that the definition of an arbitrary Cartesian product needs the Axiom of Choice to render it nonempty. --Lord_Farin 12:15, 30 November 2011 (CST)